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A 99.8% confidence interval for the average weight of a box of cereal if a sample...

A 99.8% confidence interval for the average weight of a box of cereal if a sample of 24 such boxes has an average of 18.08 ounces with a sample deviation of 0.235 ounces. The population of all such weights is normally distributed

round to the nearest hundreth of an ounce

Math   Statistics-And-Probability   
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Answer #1

X_bar =18.08

= 0.235

n=24

Df = n-1 = 24-1 =23

Now calculate 99.8% confidence interval is given by formula

CI = x_bar t*(/n)

t value for 99.8 confidence interval is with df =23 is 3.485

t = 3.485

CI = 18.08 3.485 *(0.235/24)

= 18.08 0.167

CI = 18.08 - 0.167 and CI = 18.08 + 0.167

= 17.913 CI = 18.247

Therefore 99.8% confidence interval is (17.913,18.247).

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